I would like to know how to prove that:
$$\psi^{(n)}(z)=(-1)^{n+1}n! \sum_{k=0}^{\infty}\frac{1}{\left ( k+z \right )^{n+1}}$$
I know that $\displaystyle \sum_{n=0}^{\infty}\frac{1}{n+z}=-\psi (z)$ but despite the differentiation I apply I cannot get the result . I get the alternating term but not the factorial.
For $f(z)=\frac{1}{k+z}$ you have $f'(z) = - \frac{1}{(k+z)^2}, f''(z) = - (-2) \frac{1}{(k+z)^3}, - (-2) (-3) \frac{1}{(k+z)^4}, ...$.
From $\frac{1}{k+z} = \frac{1}{k}(1-\frac{z}{k}+(\frac{z}{k})^2+(\frac{z}{k})^3-...)= \sum_{i=0}^\infty \frac{1}{i!}(\frac{1}{k+z})^{(i)}z^i$ you obtain the result ($(i)$ means i-th derivative by $z$).